N positive integer

Algebra Level 2

Find the sum of all positive integer(s) $n$ such that $n^{10} + 1$ is divisible by $n+1$ .

2 8 7 5 4 3 1 6

1 solution

Aditya Narayan Sharma
Aug 14, 2016

$\displaystyle (n+1)|(n^{10} +1)$ & also $\displaystyle (n+1)|(n^{10} -1)$ so we have $\displaystyle n+1 | (n^{10}+1)-(n^{10} -1)$

So $n+1|2$ and thus $n=1$